Q. If $H_{1}$ and $H_{2}$ be the greatest heights of a projectile in two paths for a given value of range, then the horizontal range of projectile is given by
Motion in a Plane
Solution:
$\theta_{1}+\theta_{2}=90^{\circ}$
$H_{1}=\frac{u^{2} \sin ^{2} \omega_{1}}{2 g}$
$H_{2}=\frac{u^{2} \sin \left(90^{\circ}-\theta_{1}\right)}{2 g}$
$H_{1} H_{2}=\frac{R^{2}}{16}$
$\because R=\frac{u^{2} \sin ^{2} \theta_{1}}{g}$
$R=4 \sqrt{H_{1} H_{2}}$
